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4

I'm going to use an argument along the lines of what @hOff proposed. In addition to the smoothness of $G$ I am going to suppose that it decays rapidly at infinity so that all of the following calculations are justified. For example, we could assume that $G$ is Schwartz class. As the first order of business we expand your equation as $$ \Delta G(x) + x ...


3

First, we need to show that $u \in L^n(\Omega)$. You can show by a change of variable that $$\int_\Omega |\ln \ln (1 + \frac{1}{|x|})|^n dx \leq C \int_0^1 |\ln \ln (1 + \frac{1}{r})|^n r^{n-1} dr$$ The function $|\ln \ln (1 + \frac{1}{r})|^n r^{n-1}$ is continuous on $]0,1]$, so the only problem may be on 0. But we have that for r smaller than 1/e $$1 ...


3

Necas' book "Direct Methods in the Theory of Elliptic Problems" is a wonderful guide for the topics that you mentioned above, although the proofs are very abstract and most steps are omitted. Evans' book is more understandable and also includes the topics above, but not in the most general settings of the theorems. If you just start studying the Sobolev ...


3

You're forgetting another set of eigenfunctions: $\cos(\frac{(2k+1)\pi}{2a}x)$ for $k\in\mathbb{N}$. In fact, the full set of eigenfunctions are $$\cos\left(\frac{n\pi}{2a}x\right) ~~ \text{for odd } n = 1,3,5,\ldots$$ $$\sin\left(\frac{n\pi}{2a}x\right) ~~ \text{for even } n = 2,4,6,\ldots$$ As you can see, the first eigenfunction $\cos(\pi x/2a)$ is ...


2

Since the dimension of the Euclidean space is $n$, we have $e_{n} = (0,0, \dots, 0, 1)$, where there are $n$ coordinates. Then $\lambda e_{n} = (0,0, \dots, 0, \lambda)$. Since $\Bbb R^{n}_{+} = \{(x_{1}, \dots, x_{n} \mid x_{n} > 0 \}$, then the boundary of this space is $\partial \Bbb R^{n}_{+} = \{ (x_{1}, \dots, x_{n}) \mid x_{n} = 0 \}$. Then since ...


2

As remarked in the comment, the construction is identical to the case where one minimizes $R(u)$. First of all, observe that $$\tilde R(u) = R(u) + \frac{\int_M \alpha |u|^2 }{\|u\|^2_{2}} \ge R(u) - \|\alpha\|_\infty \ge \lambda_1 - \|\alpha\|_\infty.$$ Thus $\tilde R$ is bounded below. So we can take $u_1, u_2, \cdots \in H$, so that $\|u_i\|_2 = 1$ ...


1

I will prove a different estimate, but it should help you get started and give you a feeling of how to use your knowledge. Notice that since $\eta\in C^1_0(B_R)$ and $u\in W^{1,2}(B_R)$, you have $\phi\in W^{1,2}_{B_R}$. Therefore $\phi$ is a valid test function. Note also that it follows from your ellipticity estimate that $a_{ij}X_iX_j\leq\Lambda|X||Y|$ ...


1

By Lax-Milgram there exists only one $\theta\in W_0^{1,2}$ such that \begin{equation} \int_\Omega\nabla\theta\cdot\nabla\phi dx=\int_\Omega f\phi dx \end{equation} for all $\phi\in W_0^{1,2}$. So in fact $T$ is well defined, linear and continuous from $W_0^{1,2}$ to $W_0^{1,2}$ with norm $|\theta|_{W_0^{1,2}}$. Obviously it is compact, since the range of $T$ ...


1

This PDE is separable as written. Assume solutions of the form $U(x,t)=X(x)\cdot T(t)$ $$U_{tt}+U_t=U_{xx}$$ $$XT''+XT'=X''T$$ $$\frac{T''}{T}+\frac{T'}{T}=\frac{X''}{X}=-\lambda$$ The equation separates into $$T''+T'+\lambda T=0 \mbox{ and } X''+\lambda X = 0$$ The solution to the first equation is the span of $$\{e^{-t/2}cos(\frac{\sqrt{4\lambda -1}}{2} t) ...


1

Proceed as usual. The Fourier coefficients of a piecewise defined functions are still computed as an integral: $$ a_k = \frac{1}{\pi}\int_{-\pi}^{\pi} u(1,\theta) \cos k\theta\,d\theta $$ It's just that to evaluate this, you'll have to split the integral into the sum of two: $$ a_k = \frac{1}{\pi}\int_{-\pi}^{0} u(1,\theta) \cos k\theta\,d\theta + ...


1

Recall the vector form of integration by parts in higher dimensions: $$\int_{\Omega}\nabla\phi\cdot\boldsymbol{v}\,d\Omega= \int_{\partial\Omega}\phi(\boldsymbol{v}\cdot\boldsymbol{n})\,d\Gamma-\int_{\Omega}\phi\nabla\cdot\boldsymbol{v}\,d\Omega.$$ In your case, $\boldsymbol{v} = A\nabla u$ and equality (2) follows immediately, assuming the notation $\phi ...


1

A reference I have found very useful is the book "Second Order Elliptic Equations and Elliptic Systems'' by Ya-Zhe Chen and Lan-Cheng Wu (English translation). The proofs are complete and well organized. In chapter 9, theorem 2.6 the theorem is proven when $k=\nabla \cdot F$ and $F$ is $C^{0,\alpha}$. The result also holds when $k\in L^p$ and $p>n$ and ...


1

I do not suppose this to be an answer, but comments are not permitted without login. May be you should specify which side of the inequality makes the trouble. Obviously the argument follows a standard one which I just met in on page 14 (Theorem 8.1) of http://arxiv.org/pdf/math/0309021 . Does looking at that short proof help? In the definition of $w$, I ...


1

A general way to derive a weak form is to multiply a test function on both sides of the equation and then integrate them. The second step is to use some kind of divergence theorems to derive the weak solution such that the solution is some what not so smooth as in the strong form. For your question here, we can derive the weak form as follows: Let the ...


1

It does has strict inequality: Let $f_1$ satisfies $-\Delta f_1 = \lambda f_1$ and $\|f_1\|_{L^2(\Omega_1)} = 1$ on $\Omega_1$. If $\Omega_1 \subset \Omega_2$, then the function $$g(x) = \begin{cases} f_1(x) & \text{ if } x\in \Omega_1 \\ 0 & \text{ if not.}\end{cases}$$ is a $W^{1, 2}_0(\Omega)$ function, $\|g\|_{L^2(\Omega_2)} = ...


1

Assuming you're using method of lines. Let the original initial-boundary problem be $$ u_t = u_{xx}\\ u(0, x) = f(x)\\ u(t, 0) = a(t)\\ u(t, 1) = b(t). $$ Introduce a set of points $x_j = jh,\; j = 0,1,\dots, N,\;Nh = 1$. Let $u_j(t) = u(t, x_j)$. Note that $u_0(t) = a(t),\; u_N(t) = b(t)$ are already known. Unknown are $u_j(t),\; j = 1, 2, \dots, N-1$. ...


1

My answer will be very vague. But you need to find invariants of this equation and after you find go to canonical coordinates and it will reduce the dimensionality of the equation(will make it ODE) Prof. Bluman is one of the most prominent scientist in the field of symmetries you can check out books on his website for more information. ...


1

First you can prove (by approximating $u$ and $v$ by smooth functions) that $$ D(uv) = u Dv + vDu. $$ Assume first $p<n$ and $q<n$. Then using Sobolev embedding theorem, we know that $$ u\in L^{\frac{pn}{n-p}}(\Omega), \quad v\in L^{\frac{qn}{n-q}}(\Omega). $$ Using Hoelder inequality one can prove that $f\in L^p(\Omega)$, $g\in L^q(\Omega)$ implies ...



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